3.2509 \(\int \frac{(2+3 x)^3}{\sqrt{1-2 x} (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=84 \[ -\frac{2 \sqrt{1-2 x} (3 x+2)^2}{165 (5 x+3)^{3/2}}-\frac{\sqrt{1-2 x} (9405 x+5831)}{18150 \sqrt{5 x+3}}+\frac{81 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{50 \sqrt{10}} \]

[Out]

(-2*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(165*(3 + 5*x)^(3/2)) - (Sqrt[1 - 2*x]*(5831 + 9405*x))/(18150*Sqrt[3 + 5*x]) +
(81*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(50*Sqrt[10])

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Rubi [A]  time = 0.0191753, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {98, 143, 54, 216} \[ -\frac{2 \sqrt{1-2 x} (3 x+2)^2}{165 (5 x+3)^{3/2}}-\frac{\sqrt{1-2 x} (9405 x+5831)}{18150 \sqrt{5 x+3}}+\frac{81 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{50 \sqrt{10}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x)^3/(Sqrt[1 - 2*x]*(3 + 5*x)^(5/2)),x]

[Out]

(-2*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(165*(3 + 5*x)^(3/2)) - (Sqrt[1 - 2*x]*(5831 + 9405*x))/(18150*Sqrt[3 + 5*x]) +
(81*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(50*Sqrt[10])

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 143

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x
)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)), x] + Dist[(a*d*f*h*m + b*(d*(f*g + e*h) - c*f*h*(m +
2)))/(b^2*d), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m
+ n + 2, 0] && NeQ[m, -1] &&  !(SumSimplerQ[n, 1] &&  !SumSimplerQ[m, 1])

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{(2+3 x)^3}{\sqrt{1-2 x} (3+5 x)^{5/2}} \, dx &=-\frac{2 \sqrt{1-2 x} (2+3 x)^2}{165 (3+5 x)^{3/2}}-\frac{2}{165} \int \frac{\left (-109-\frac{285 x}{2}\right ) (2+3 x)}{\sqrt{1-2 x} (3+5 x)^{3/2}} \, dx\\ &=-\frac{2 \sqrt{1-2 x} (2+3 x)^2}{165 (3+5 x)^{3/2}}-\frac{\sqrt{1-2 x} (5831+9405 x)}{18150 \sqrt{3+5 x}}+\frac{81}{100} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{2 \sqrt{1-2 x} (2+3 x)^2}{165 (3+5 x)^{3/2}}-\frac{\sqrt{1-2 x} (5831+9405 x)}{18150 \sqrt{3+5 x}}+\frac{81 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{50 \sqrt{5}}\\ &=-\frac{2 \sqrt{1-2 x} (2+3 x)^2}{165 (3+5 x)^{3/2}}-\frac{\sqrt{1-2 x} (5831+9405 x)}{18150 \sqrt{3+5 x}}+\frac{81 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )}{50 \sqrt{10}}\\ \end{align*}

Mathematica [A]  time = 0.0458114, size = 60, normalized size = 0.71 \[ -\frac{\sqrt{1-2 x} \left (49005 x^2+60010 x+18373\right )}{18150 (5 x+3)^{3/2}}-\frac{81 \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{50 \sqrt{10}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x)^3/(Sqrt[1 - 2*x]*(3 + 5*x)^(5/2)),x]

[Out]

-(Sqrt[1 - 2*x]*(18373 + 60010*x + 49005*x^2))/(18150*(3 + 5*x)^(3/2)) - (81*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]])
/(50*Sqrt[10])

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Maple [A]  time = 0.012, size = 113, normalized size = 1.4 \begin{align*}{\frac{1}{363000} \left ( 735075\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}+882090\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x-980100\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}+264627\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) -1200200\,x\sqrt{-10\,{x}^{2}-x+3}-367460\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{1-2\,x}{\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}} \left ( 3+5\,x \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+3*x)^3/(3+5*x)^(5/2)/(1-2*x)^(1/2),x)

[Out]

1/363000*(735075*10^(1/2)*arcsin(20/11*x+1/11)*x^2+882090*10^(1/2)*arcsin(20/11*x+1/11)*x-980100*x^2*(-10*x^2-
x+3)^(1/2)+264627*10^(1/2)*arcsin(20/11*x+1/11)-1200200*x*(-10*x^2-x+3)^(1/2)-367460*(-10*x^2-x+3)^(1/2))*(1-2
*x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)

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Maxima [A]  time = 3.71051, size = 103, normalized size = 1.23 \begin{align*} \frac{81}{1000} \, \sqrt{5} \sqrt{2} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) - \frac{27}{250} \, \sqrt{-10 \, x^{2} - x + 3} - \frac{2 \, \sqrt{-10 \, x^{2} - x + 3}}{4125 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac{602 \, \sqrt{-10 \, x^{2} - x + 3}}{45375 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^3/(3+5*x)^(5/2)/(1-2*x)^(1/2),x, algorithm="maxima")

[Out]

81/1000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 27/250*sqrt(-10*x^2 - x + 3) - 2/4125*sqrt(-10*x^2 - x + 3)/(
25*x^2 + 30*x + 9) - 602/45375*sqrt(-10*x^2 - x + 3)/(5*x + 3)

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Fricas [A]  time = 1.78424, size = 285, normalized size = 3.39 \begin{align*} -\frac{29403 \, \sqrt{10}{\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac{\sqrt{10}{\left (20 \, x + 1\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{20 \,{\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \,{\left (49005 \, x^{2} + 60010 \, x + 18373\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{363000 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^3/(3+5*x)^(5/2)/(1-2*x)^(1/2),x, algorithm="fricas")

[Out]

-1/363000*(29403*sqrt(10)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10
*x^2 + x - 3)) + 20*(49005*x^2 + 60010*x + 18373)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (3 x + 2\right )^{3}}{\sqrt{1 - 2 x} \left (5 x + 3\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)**3/(3+5*x)**(5/2)/(1-2*x)**(1/2),x)

[Out]

Integral((3*x + 2)**3/(sqrt(1 - 2*x)*(5*x + 3)**(5/2)), x)

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Giac [B]  time = 2.20475, size = 220, normalized size = 2.62 \begin{align*} -\frac{\sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}}{3630000 \,{\left (5 \, x + 3\right )}^{\frac{3}{2}}} - \frac{27}{1250} \, \sqrt{5} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5} + \frac{81}{500} \, \sqrt{10} \arcsin \left (\frac{1}{11} \, \sqrt{22} \sqrt{5 \, x + 3}\right ) - \frac{201 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}{302500 \, \sqrt{5 \, x + 3}} + \frac{{\left (\frac{603 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} + 4 \, \sqrt{10}\right )}{\left (5 \, x + 3\right )}^{\frac{3}{2}}}{226875 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^3/(3+5*x)^(5/2)/(1-2*x)^(1/2),x, algorithm="giac")

[Out]

-1/3630000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) - 27/1250*sqrt(5)*sqrt(5*x + 3)*sqr
t(-10*x + 5) + 81/500*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 201/302500*sqrt(10)*(sqrt(2)*sqrt(-10*x +
5) - sqrt(22))/sqrt(5*x + 3) + 1/226875*(603*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4*sq
rt(10))*(5*x + 3)^(3/2)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3